You are given the array nums consisting of n positive integers. You computed the sum of all non-empty continuous subarrays from the array and then sorted them in non-decreasing order, creating a new array of n * (n + 1) / 2 numbers.
Return the sum of the numbers from index left to index right (indexed from 1), inclusive, in the new array. Since the answer can be a huge number return it modulo 109 + 7.
Example 1:
Input: nums = [1,2,3,4], n = 4, left = 1, right = 5 Output: 13 Explanation: All subarray sums are 1, 3, 6, 10, 2, 5, 9, 3, 7, 4. After sorting them in non-decreasing order we have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 1 to ri = 5 is 1 + 2 + 3 + 3 + 4 = 13.
Example 2:
Input: nums = [1,2,3,4], n = 4, left = 3, right = 4 Output: 6 Explanation: The given array is the same as example 1. We have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 3 to ri = 4 is 3 + 3 = 6.
Example 3:
Input: nums = [1,2,3,4], n = 4, left = 1, right = 10 Output: 50
Constraints:
n == nums.length1 <= nums.length <= 10001 <= nums[i] <= 1001 <= left <= right <= n * (n + 1) / 2class Solution {
public int rangeSum(int[] nums, int n, int left, int right) {
int idx = 0;
int[] arr = new int[n * (n + 1) / 2];
arr[0] = nums[0];
for (int i = 1; i < n; i++)
arr[++idx] = nums[i] += nums[i - 1];
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
arr[++idx] = nums[j] - nums[i]; // subarray sum
// System.out.println(Arrays.toString(arr));
Arrays.sort(arr);
// System.out.println(Arrays.toString(arr));
int mod = (int) 1e9 + 7, sum = 0;
for (int i = left - 1; i < right; i++)
sum = (sum + arr[i]) % mod;
return sum;
}
}